Given L=limn→∞n∑r=1(1r!r∏i=1(i2+13)). then ⌊L⌋ is
Try: limn→∞n∑r=11r![(12+13)⋅(22+13)⋯⋯(r2+13)]
could some help me to solve it, thanks
Answer
For any |a|<1,∞∑n=11n!n∏k=1(ak+b)=∞∑n=1(−a)nn!n∏k=1(−ba−k)=∞∑n=1(−a)nn!(−ba−1n).
Note that |a|<1, by the generalized binomial theorem,(1−a)−ba−1=∞∑n=0(−a)nn!(−ba−1n),
thus∞∑n=11n!n∏k=1(ak+b)=∞∑n=0(−a)nn!(−ba−1n)−1=(1−a)−ba−1−1.
In this question, take a=12 and b=13, thenL=∞∑n=11n!n∏k=1(k2+13)=253−1,
and [L]=2.
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